finite boolean ring order is $2^n$

Question

let $R$ be a finite boolean ring.
prove that $|R|=2^n$ for some $n\in\mathbb N$.

I know that $R$ is commutative and for every element $a\in R\space a+a=0$ and $a^2=a$

Answer 1

Cauchy's theorem -- given a group $G$ and a prime $p$ such that $p$ divides $|G|$, there exists an element of $G$ of order $p$. Considering $(R,+)$ as an abelian group, this should give you a proof. (You actually only need the abelian case of Cauchy's theorem for this problem, which is quite easy to prove).

Answer 2

A boolean ring is an algebra, in particular a vector space, over the two element field $\{0,1\}$.

Alternatively, the additive group of $R$ is a $2$-group: the order of every element is a power of $2$. A finite $p$-group ($p$ a prime) has order $p^n$ for some integer $n$.